Inspired by nam le - Part 1

Algebra Level 3

Let $a,b,c$ be non-negative reals so that $a+b+c=1$ . If the value of $\sqrt{16a+9}+\sqrt{16b+9}+\sqrt{16c+9}$ lies on the interval [ $x,y$ ], find the value of $x^2+y^2$ .

Inspiration.

The answer is 250.

1 solution

Zach Bian
Dec 29, 2017

We know that $y=\sqrt{16x+9}$ is an increasing function, concave down, which means for a higher x, you get diminishing returns.

First, find the upper bound. Evenly raising a, b, and c gives the highest value, so they must be equal, which means they are all $\frac{1}{3}$ . Raising a, b, or c past $\frac{1}{3}$ , means at least one of the others has to go down to compensate, which means the sum goes down. The $y^2$ is therefore 129.

Then, to find the lower bound, make a, b, or c equal 1, and the others equal 0. Lowering that value causes at least one of the others to go up, which means a higher sum. So $x^2$ is 121.

121+129 = 250.

Inspired by nam le - Part 1

The answer is 250.

1 solution

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